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Integration of rational functions

  1. Mar 6, 2004 #1
    How do I solve the integration of a rational function such as:

    x^2 - 6x - 2
    (x^2 + 2)^2

    If possible, please list the general rule of solving, I DO NOT want the answer, I simply want to know the way of solving it.
    Thanks in advance!

    So far I got to the part where A = 1, B = -6 and C = -4, what can I do next?
     
  2. jcsd
  3. Mar 6, 2004 #2

    Hurkyl

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    Well, I guess that's one good thing about standardized notation across calc texts; when the student gives absolutely no indication about what they're doing, you can still tell from the variables they used!

    I'm assuming what you've done already is partial fractions.

    Anyways, the integrals you're left with should looke like the type of integral you've done in previous sections. (you spent an entire section on them!) You need to make a substitution...



    In general, you might need to complete the square in the quadratic factors in the denominator to get them in the right form.
     
  4. Mar 6, 2004 #3
    I am new here and I don't how to do fractions so ignore the
    ............ I need it for the space


    You have shown
    x^2 - 6x - 2

    (x^2 + 2)^2

    = 1 ................ -6x-4
    ------- +.. ----------
    (x^2+2) .... (x^2+2)^2

    The first fraction integral is a standard integral ( inverse tan)

    and a hint is the first fraction can be rewritten as

    -3*(2x) ............... -4
    ----------- + .. --------
    (x^2+2)^2 ..... (x^2+2)^2


    Use the subsitution u =x^2+2 for the first fraction
    and I think the substitution u = square root of 2 * tan u for the
    second fraction.
     
    Last edited: Mar 6, 2004
  5. Mar 6, 2004 #4
    The problem is,I am in grade 12 and I have never done integration before (In fact, we are doing derivative right now)

    I am trying to check the derivative:
    x^2 - 6x - 2
    (x^2 + 2)^2
    that I got from:
    -x + 3
    x^2 + 2

    So I figure the anti-derivative (Integral) might be able to help me check, who would have thought it is so much work? I should have pick graphing instead.:smile:

    I haven't got a clue how to make the derivative looks like original function shown above by integration, and althought i have not done any integration, I would like to still give it a try. Any help would be welcome, thanks!:smile:
     
    Last edited: Mar 6, 2004
  6. Mar 6, 2004 #5

    Hurkyl

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    There are two things you need to look for in your text: "Substitution" and "Trigonometric substitution". Everything you need to integrate that (now that you've done "Partial Fractions") will be spelled out for you, probably much more thoroughly than you can get from a post on a message board.


    That being said, there are other ways you can check a derivative. You can try plotting your original function, and then check to make sure your derivative is positive whenever the original function is increasing.

    Rolle's theorem says that if a function is zero at two different points, the derivative has to be zero someplace in-between; so you could look for the zeroes of both.

    You could try making a differential approximation. Remember that:

    [tex]
    f(x + \delta x) \approxeq f(x) + f'(x) \delta x
    [/tex]

    is a good approximation when δx is small, so you could plug in some values and check. (The error is eventually much smaller than δx)
     
    Last edited: Mar 6, 2004
  7. Mar 6, 2004 #6
    Integration can be much harder than differentiation. One reason is that you do not have something as easy to use the product rule, quotient rule etc. (you can work backwards using integration by substitution but)

    to make this clear even though though

    1
    ----
    1+x^2

    is a rational function it integrated to inverse tan x which is not one polynomial divided by another.
     
  8. Mar 6, 2004 #7
    I got something like:

    6x + 4 + In |x^2 + 2| + C
    x^2 + 2

    which is nothing like:
    -x + 3
    x^2 + 2
     
  9. Mar 7, 2004 #8

    Hurkyl

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    And in all truth, I think that antidifferentiation here is far more error prone that differentiation. :smile: You could probably compute the derivative several times (to make sure there are no arithmetic mistakes!) in the time it takes to perform the antidifferentiation, and you probably have smaller odds of making a mistake too. :smile:
     
  10. Mar 7, 2004 #9

    NateTG

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    I think Hurkyl is right on the money with his notion of numeric approximation of the derivative. It's a reasonable way to do a santiy check on the work.

    That said, when i differentiate:
    [tex]\frac{d}{dx} \frac{-x + 3}{x^2+2}[/tex]
    I get
    [tex]\frac{-1}{x^2+2} - 2x \frac{-x+3}{x^4+4x^2+4}[/tex]
    or
    [tex]\frac{x^2-6x-2}{x^4+4x^2+4}[/tex]

    which looks similar to your answer.
     
    Last edited: Mar 7, 2004
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